This example comes from the book by Elias Wegert: Visual Complex Functions. Consider the function $f(z):=z^{1/2}$ for $z \in \mathbb{C}$ with $|z – 1| < 1.$ For these $z$, the function can be represented by a convergent power series
$$f(z) = \sum_{k=0}^{\infty} {1/2 \choose k} (z-1)^{k},$$
which means that $z^{1/2}$ is analytic in $D(1,1).$ Now, Theorem 3.3.1 in that book says that, given
$$f(z) = \sum_{k=0}^{\infty} a_k(z-z_0)^k, \quad \forall z \in D(z_0,r),$$
you can "rearrange" the latter series into
$$f(z) = \sum_{k=0}^{\infty} b_k (z-z_1)^k, \quad \forall z \in D(z_1,r_1),$$
where the $b_{k}$'s are given by $\sum_{n=k}^{\infty} {n \choose k} a_n (z_1-z_0)^{n-k},$ $k = 0, 1, \ldots, $ provided only that $|z_1 – z_0| < r$ and $r_1 := r – |z_1 – z_0|.$ Noting that $\rm{e}^{i \pi/4} \in D(1,1)$ we can try to apply the Theorem to the series defining $z^{1/2}$ with $z_1 = \rm{e}^{i \pi/4}$. This should give
$$f(z) = \sum_{k=0}^{\infty} \left(\sum_{n=k}^{\infty} {n \choose k} {1/2 \choose n} (\rm{e}^{i \pi/4} – 1)^{n-k} \right) (z-z_1)^k.$$
This in turn should be equal to
$$f(z) = \rm{e}^{i \pi/8} \sum_{k=0}^{\infty}\rm{e}^{-i k\pi/4} {1/2 \choose k} (z-z_1)^k,$$
the point being here that while the original series diverges at, say, $z = i,$ the last one is already convergent there (…). The only thing that completely eludes me is why (as it seems)
$$\sum_{n=k}^{\infty} {n \choose k} {1/2 \choose n} (\rm{e}^{i \pi/4} – 1)^{n-k} = \rm{e}^{i \pi/8}\rm{e}^{-i k\pi/4} {1/2 \choose k},$$
for $k = 0, 1 \ldots.$ I thought about induction on $k$ but was only able to do the base step $k=0.$ Could anybody help me? Thanks.
Best Answer
Remember that the coefficients of the Taylor series of $f$ about $z_1$ are $\dfrac{f^{(k)}(z_1)}{k!}$. Writing $z^{1/2}$ for the square root, we have
$$z_1^{1/2} = \sum_{n=0}^\infty \binom{1/2}{n}(z_1-1)^n,$$
which is the $k= 0$ case. Now differentiate the identity with respect to $z_1$,
$$\frac{1}{2}z_1^{1/2}z_1^{-1} = \sum_{n=1}^\infty \binom{n}{1}\binom{1/2}{n}(z_1-1)^{n-1}$$
is the case $k = 1$. Generally,
$$\frac{d}{dz_1} z_1^{1/2-k}\binom{1/2}{k} = (k+1)\cdot z_1^{1/2-(k+1)}\binom{1/2}{k+1},$$
and
$$\begin{align} \frac{d}{dz_1}\sum_{n=k}^\infty \binom{n}{k}\binom{1/2}{n}(z_1-1)^{n-k} &= \sum_{n=k+1}^\infty (n-k)\binom{n}{k}\binom{1/2}{n}(z_1-1)^{n-(k+1)}\\ &= \sum_{n=k+1}^\infty (k+1)\binom{n}{k+1}\binom{1/2}{n}(z_1-1)^{n-(k+1)}. \end{align}$$
Divide by $k+1$ to obtain the $k+1$ case from the $k$ case.
Nicer, in my opinion, is a direct exploitation of the series centered at $1$: Near $z_1$, where $\lvert z_1-1\rvert < 1$, we have
$$\begin{align} \sqrt{z} &= \sqrt{z_1}\cdot \sqrt{\frac{z}{z_1}}\\ &= \sqrt{z_1}\sqrt{1 + \left(\frac{z}{z_1}-1\right)}\\ &= \sqrt{z_1}\sum_{k=0}^\infty \binom{1/2}{k}\left(\frac{z}{z_1}-1\right)^k\\ &= \sqrt{z_1}\sum_{k=0}^\infty z_1^{-k}\binom{1/2}{k}(z-z_1)^k. \end{align}$$